Question

A tangent $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the axes at $A$ and $B$. Then the locus of mid point of $AB$ is

• A. $\frac{a^2}{x^2}+\frac{b^2}{y^2}=2$
• B. $\frac{a^2}{x^2}+\frac{b^2}{y^2}=4$
• C. $\frac{a^2}{x^2}+\frac{b^2}{y^2}=1$
• D. $\frac{a^2}{x^2}+\frac{b^2}{y^2}=\frac{1}{2}$

Answer 1

The correct option is B $\frac{a^2}{x^2}+\frac{b^2}{y^2}=4$

Tangent to the ellipse is given by,

$\frac{xcosm\theta }{a}+\frac{ysin\theta }{b}=1$

Since this line intersect the cordinate axis

at $y=0,\frac{xcos\theta }{a}=1⟹x=\frac{a}{cos\theta }$

so the cordinate is $(\frac{a}{cos\theta },0)$

Similarly

at $x=0,\frac{ysin\theta }{b}=1⟹y=\frac{b}{sin\theta }$

Cordinate is $(0,\frac{b}{sin\theta })$

Midpoint is given by-

$(h,k)=\frac{\frac{a}{cos\theta }+0}{2},\frac{0+\frac{b}{sin\theta }}{2}$

$(h,k)=(\frac{a}{2cos\theta },\frac{b}{2sin\theta })$

we get.

$cos\theta =\frac{a}{3h},sin\theta =\frac{b}{2k}$

As we know

$si{n}^2\theta +co{s}^2\theta =1⟹\frac{a^2}{4h^2}+\frac{b^2}{4k^2}=1$

$\frac{a^2}{h^2}+\frac{b^2}{k^2}=4$

therefore the locus of midpoint is

$\frac{a^2}{x^2}+\frac{b^2}{y^2}=4$